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practice: momentum math 1. find the momentum of a 25 kg object travelin…

Question

practice: momentum math

  1. find the momentum of a 25 kg object traveling at a speed of 4 m/s
  2. a ball has 2 kg*m/s of momentum when thrown with a velocity of 5 m/s outwards. find the mass of the ball.
  3. a 25 kg cart has 120 kg*m/s of momentum. how fast is the car going?
  4. a professional baseball player can pitch a baseball with a velocity of 44.7 m/s towards home plate. if a baseball weighs 1.4 n, how much momentum does it have when it hits the catchers mitt?
  5. challenge: in a collision, a 15 kg object moving with a velocity of 3 m/s transfers all of its momentum to a 5 kg object. what would be the velocity of the 5 kg object after the collision? (hint: think about the law of conservation of momentum and draw a picture of what is happening to get started!)

Explanation:

Step1: Recall the formula for momentum

The formula for momentum \( p = mv \), where \( m \) is mass and \( v \) is velocity.

Step2: Substitute values for problem 1

Given \( m = 25\space kg \) and \( v = 4\space m/s \). Then \( p=(25)(4) = 100\space kg\cdot m/s \).

Step3: Rearrange formula for problem 2

From \( p = mv \), we can solve for \( m=\frac{p}{v} \). Given \( p = 2\space kg\cdot m/s \) and \( v = 5\space m/s \). Then \( m=\frac{2}{5}=0.4\space kg \).

Step4: Rearrange formula for problem 3

From \( p = mv \), solve for \( v=\frac{p}{m} \). Given \( p = 120\space kg\cdot m/s \) and \( m = 25\space kg \). Then \( v=\frac{120}{25}=4.8\space m/s \).

Step5: Find mass for problem 4

Weight \( W = mg \), so \( m=\frac{W}{g} \). Taking \( g = 9.8\space m/s^{2} \) and \( W = 1.4\space N \). Then \( m=\frac{1.4}{9.8}\approx0.143\space kg \). Using \( p = mv \) with \( v = 44.7\space m/s \), \( p=(0.143)(44.7)\approx6.39\space kg\cdot m/s \).

Step6: Apply conservation of momentum for problem 5

By the law of conservation of momentum \( m_1v_1=m_2v_2 \). Given \( m_1 = 15\space kg \), \( v_1 = 3\space m/s \), \( m_2 = 5\space kg \). Then \( v_2=\frac{m_1v_1}{m_2}=\frac{(15)(3)}{5}=9\space m/s \).

Answer:

  1. \( 100\space kg\cdot m/s \)
  2. \( 0.4\space kg \)
  3. \( 4.8\space m/s \)
  4. \( \approx6.39\space kg\cdot m/s \)
  5. \( 9\space m/s \)